Playground
A charged particle curves in a magnetic field. Adjust B, velocity, and charge to see the circular orbit change.
Variables
| Symbol | Name | SI | Dimension | Range |
|---|---|---|---|---|
| Lorentz forceoutput Total electromagnetic force on the charged particle | N | M·L·T⁻² | 0 – 1e-10 | |
| Charge Electric charge of the particle | C | I·T | -1e-17 – 1e-17 | |
| Velocity Speed of the charged particle | m/s | L·T⁻¹ | 0 – 299792458 | |
| Magnetic field External magnetic field strength | T | M·T⁻²·I⁻¹ | 0 – 10 | |
| Electric field External electric field strength | V/m | M·L·T⁻³·I⁻¹ | 0 – 1000000 |
Deep dive
Derivation
Postulated by Lorentz as the force on a charge in given E and B fields. Consistent with special relativity: the electric and magnetic forces transform into each other under Lorentz boosts. The magnetic force is a relativistic correction to the electric force between moving charges.
Experimental verification
Thomson's cathode ray experiments (1897) measured e/m using crossed E and B fields. Cyclotron resonance, mass spectrometry, and Hall effect measurements all confirm the Lorentz force to high precision.
Common misconceptions
- The magnetic force does NO work — it changes direction but not speed
- The force is perpendicular to v, not to B (it's perpendicular to both via the cross product)
- The 'magnetic force' between two wires is really a relativistic electric effect in the electrons' rest frame
Real-world applications
- Cyclotrons and synchrotrons in particle physics research
- Mass spectrometers for chemical and isotope analysis
- Hall effect sensors in automotive and industrial applications
- Aurora borealis — charged solar wind particles spiral along Earth's magnetic field
Worked examples
Cyclotron radius of an electron
Given:
- v:
- 2000000
- B:
- 0.01
- m:
- 9.1093837015e-31
- q:
- 1.602176634e-19
Find: Radius of circular orbit
Solution
r = mv/(qB) = 9.109×10⁻³¹ × 2×10⁶ / (1.602×10⁻¹⁹ × 0.01) = 1.138×10⁻³ m ≈ 1.14 mm
Velocity selector (crossed E and B fields)
Given:
- E_field:
- 100000
- B:
- 0.05
Find: Selected velocity
Solution
v = E/B = 10⁵ / 0.05 = 2×10⁶ m/s
Scenarios
What if…
- scenario:
- What if the particle is a proton instead of an electron?
- answer:
- r = m_p × v/(qB) = 1.673×10⁻²⁷ × 2×10⁶ / (1.602×10⁻¹⁹ × 0.01) = 2.09 m — about 1836× larger than the electron orbit.
- scenario:
- What if the velocity is parallel to B?
- answer:
- No magnetic force (cross product = 0). The particle travels in a straight line along B — no deflection at all.
- scenario:
- What if both E and B are present and perpendicular?
- answer:
- Particle drifts perpendicular to both E and B with velocity v_drift = E×B/B². This is the E×B drift, fundamental in plasma physics.
Limiting cases
- condition:
- v = 0
- result:
- F = qE (electric force only)
- explanation:
- A stationary charge feels no magnetic force — only electric fields act on it.
- condition:
- E = 0
- result:
- F = qvB sin θ
- explanation:
- Pure magnetic force — always perpendicular to velocity, so it does no work.
- condition:
- v ∥ B
- result:
- F_magnetic = 0
- explanation:
- Cross product vanishes for parallel vectors — the particle moves in a straight line along B.
Context
Hendrik Lorentz · 1895
Lorentz unified the electric and magnetic forces on a moving charge into a single expression, providing the bridge between Maxwell's field equations and Newton's mechanics of particles.
Hook
Why do charged particles spiral in the Northern Lights instead of flying straight?
An electron moving at 2 × 10⁶ m/s enters a 0.01 T magnetic field perpendicular to its velocity. Find the radius of its circular orbit.
Dimensions: [F] = [q]([E] + [v]·[B]) → C(V/m + m/s·T) = C(N/C + N/C) = N ✓
Validity: Valid for point charges at all speeds when combined with relativistic momentum. The non-relativistic form (F = ma) applies for v << c. Does not account for radiation reaction — for that, use the Abraham–Lorentz force.